See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: SNAr (nucleophilic aromatic substitution) proceeds via a Meisenheimer complex intermediate. The rate-determining step is the addition of the nucleophile to form this anionic intermediate. Electron-withdrawing groups (EWGs) ortho and para to the leaving group stabilize the negative charge in the Meisenheimer complex and thus increase the rate. Meta EWGs provide inductive stabilization but do NOT directly delocalize the negative charge into the ring positions bearing the substituents, making them less effective activators. Step 2 - Identify each compound: (i) 1-chloro-4-nitrobenzene: one NO2 para to Cl — activates via resonance delocalization at ortho/para positions relative to Cl (para NO2 directly stabilizes the Meisenheimer complex). (ii) 1-chloro-3-nitrobenzene: one NO2 meta to Cl — activates only inductively, not by resonance at the ipso/ortho/para positions relative to Cl. (iii) 2,4-dinitrochlorobenzene: two NO2 groups (ortho and para to Cl) — both directly stabilize the Meisenheimer complex by resonance. Greater activation than one NO2. (iv) Chlorobenzene: no activating EWG — very slow SNAr, essentially unreactive under normal conditions. (v) 2,4,6-trinitrochlorobenzene (picryl chloride): three NO2 groups (two ortho and one para to Cl) — maximum resonance stabilization of the Meisenheimer complex. Fastest SNAr rate. Step 3 - Rank by rate: - (v) has three NO2 groups at ortho/ortho/para: highest activation → fastest rate. - (iii) has two NO2 groups at ortho and para: second highest. - (i) has one NO2 at para: third. - (ii) has one NO2 at meta: only inductive effect, no resonance stabilization of Meisenheimer complex → slower than para isomer. - (iv) has no EWG: slowest. Decreasing order: v > iii > i > ii > iv, which corresponds to option (c). Step 4 - Why other options fail: (a) i > ii > iv > iii > v — incorrect; places compounds with more NO2 groups lower. (b) ii > i > iii > v > iv — incorrect; meta isomer cannot be faster than para isomer. (d) v > iii > ii > i > iv — incorrect; places meta (ii) above para (i), which contradicts resonance stabilization principles. Therefore, the correct answer is C.